Problem: $\overline{AC} = 5$ $\overline{BC} = {?}$ $A$ $C$ $B$ $5$ $?$ $ \sin( \angle BAC ) = \frac{6\sqrt{61} }{61}, \cos( \angle BAC ) = \frac{5\sqrt{61} }{61}, \tan( \angle BAC ) = \dfrac{6}{5}$
Answer: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{\overline{BC}}{5} $ $ \overline{BC}=5 \cdot \tan( \angle BAC ) = 5 \cdot \dfrac{6}{5} = 6$